Q.
From a circular disc of radius $R$ and mass 9 $M$, a small disc of radius $R$/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through $O$ is
A) $4\ {MR}^{2}$
B) $\frac{40}{9}\ {MR}^{2}$
C) $10\ {MR}^{2}$
D) $\frac{37}{9}\ {MR}^{2}$
Explanation
(a) $I_{remaining} = I_{whole} - I_{removed}$
or $I = \frac{1}{2}(9M)\left( R^{2} \right) - \left\lbrack \frac{1}{2}m\left( \frac{R}{3} \right)^{2} + \frac{1}{2}m\left( \frac{2R}{3} \right)^{2} \right\rbrack$ …(i)
Here, $m = \frac{9M}{\pi R^{2}} \times \pi\left( \frac{R}{3} \right)^{2} = M$
Substituting in Eq. (i), we have
$I = 4MR^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.